Subject: Re: [xsl] xsl transform From: "Joris Gillis" <roac@xxxxxxxxxx> Date: Tue, 28 Jun 2005 16:33:35 +0200 |
<xsl:template match="row"> (this I don't know how to make it work) <e> <xsl:for-each select="*"> <xsl:attribute name="????"> <xsl:value-of select="."/> </xsl:attribute> </xsl:for-each> </e> </xsl:template>
regards, -- Joris Gillis (http://users.telenet.be/root-jg/me.html) "N&N9N;N.N:N?N?N= N5N9N=N1N9 N<N1N;N;N?N= N7 ON9N;ON;N1N;N?N=" - NN;N5ON2N?ON;N?O
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